Chemistry

Thermochemistry and Equilibrium

121 Questions

Thermochemistry and equilibrium problems focus on calculating bond energies, lattice enthalpy, and the Born Haber cycle. These concepts are vital for scoring well in chemistry sections. Regular practice ensures a clear understanding of energy changes in reactions.

Bond energy calculationsBorn Haber cycleLattice enthalpyEnthalpy of solutionThermochemical equations

Thermochemistry and Equilibrium Questions

Multiple choice chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

Enthalpy of the system is given as :

  1. $\displaystyle H+PV$
  2. $\displaystyle U+PV$
  3. $\displaystyle U-PV$
  4. $\displaystyle H-PV$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Enthalpy: Chemical reactions are generally carried out at constant pressure (atmospheric pressure) so it has been found useful to define a new state function Enthalpy $(H)$ as :
$H = U + PV$ 

Multiple choice chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

Enthalpy of the system is given as

  1. $\,H + PV$
  2. $\,U + PV$
  3. $\,U - PV$
  4. $\,H - PV$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Chemical reactions are generally carried out at constant pressure (atmospheric pressure) so it has been found useful to define a new state function Enthalpy (H) as :
$H=U+PV $ (By definition)
$\Delta H=\Delta U+\Delta (PV)$
$\Delta H=\Delta U+P\Delta V$ (at constant pressure) combining with first law.
$\Delta H=q _{p}$
Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

If the heat of combustion of carbon monoxide at constant volume and at $17^o$C is $-283.3$ kJ, then its enthalpy of combustion at constant pressure($R=8.314J degree^{-1} mol^{-}$)

  1. $-284.5$ kJ
  2. $284.5$ kJ
  3. $384.5$ kJ
  4. $-384.5$ kJ
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Solution:- (A) $- 284.5 \; kJ$
Heat change at constant volume for the combustion of carbon monoxide $= -283.3 \; kJ$
${CO} _{\left( g \right)} + \cfrac{1}{2} {{O} _{2}} _{\left( g \right)} \longrightarrow {C{O} _{2}} _{\left( g \right)}$
From the above reaction,
$\Delta{{n} _{g}} = {n} _{P} - {n} _{R} = 1 - \left( \cfrac{1}{2} + 1 \right) = - \cfrac{1}{2}$
Temperature $\left( T \right) = 17 ℃ = \left( 17 + 273 \right) K = 290 K \; \left( \text{Given} \right)$
Now from first law of thermodynamics,
$\Delta{H} = \Delta{E} + \Delta{{n} _{g}} RT$
$\Delta{H} = -283.3 + \left( -\cfrac{1}{2} \right) \times 8.314 \times {10}^{-3} \times 290$
$\Rightarrow \Delta{H} = -283.3 - 1.205 = - 284.505 \; kJ$
Hence the heat of reaction at constant pressure will be $- 284.5 \; kJ$.
Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

A sample of $CH _4$ of 0.08 g was subjected to combustion at $27^oC$ in a bomb calorimeter. The temperature of the calorimeter system was found to be raised by $0.25^oC$. If heat capacity of calorimeter is 18 kJ, $\Delta H$ for combustion of $CH _4$ at $27^oC$ is:

  1. $- 900$ kJ/mole
  2. $- 905$ kJ/mole
  3. $- 895$ kJ/mole
  4. $- 890$ kJ/mole
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${ CH } _{ 4 }\left( g \right) +{ 2O } _{ 2 }\left( g \right) \longrightarrow { CO } _{ 2 }\left( g \right) +2{ H } _{ 2 }O\left( l \right) $

$\Delta E=$ Heat of combustion
        $=$ Heat capacity $\times$ rise in $T$ $\times$ $\dfrac { Molar\ mass }{ Mass\ of\ compound } $
        $=-18\times 0.25\times \dfrac { 16 }{ 0.08 } $
        $=-900$ KJ/mole                                         $R=8.314\times { 10 }^{ -3 }KJ{ mol }^{ -1 }$
$\Delta H=\Delta E+\Delta nRT$                                      $\Delta n=1-3=-2$
$=-900+\left( -2 \right) \times 8.314\times { 10 }^{ -3 }\times 300$           $T=300K$
$=-900-4.9884=-904.98\simeq -905KJ/mol$

Multiple choice physics energy : forms and sources concept of energy energy for everything forms of energy

Energy equivalent of 4.2 mg in kilo calories is

  1. $9 \times 10^{10} cal$
  2. $8 \times 10^{8} cal$
  3. $10 \times 10^{9} cal$
  4. $6 \times 10^{7} cal$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using E = mc^2. m = 4.2 mg = 4.2 * 10^-6 kg. c = 3 * 10^8 m/s. E = 4.2 * 10^-6 * (9 * 10^16) = 37.8 * 10^10 J. Converting to calories (1 cal = 4.184 J), the result is approximately 9 * 10^10 cal.

Multiple choice chemistry the world of carbon organic compounds allotropes of carbon versatile nature of carbon

$C _{graphite}+O _{2(g)}\rightarrow CO _{2(g)}; \Delta H=-393.5kJ, \Delta H$ of the reaction cannot be:

  1. Heat of formation of $CO _{2}$
  2. Heat of combustion of $CO _{2}$

  3. Heat of reaction

  4. Heat of transition
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$C(graphite)+O _2(g)\longrightarrow CO _2(g)$


Now, here $\Delta H$ of the reaction cannot be the heat of transition, because there is no transition between phase take place at constant temperature and pressure.

For example: transition of $C _{diamond}\to C _{graphite}$

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

The enthalpy of tetramerization of $X$ in gas phase $(4X(g)\rightarrow { X } _{ 4 }(g))$ is $-100\ kJ/mol$ at $300\ K$. The enthalpy of vaporisation for liquid $X$ and ${X} _{4}$ are respectively $30\ kJ/mol$ and $72\ kJ/mol$ respectively.
$\Delta S$ for tetramerization of $X$ in liquid phase is $-125\ J/K mol$ at $300\ K$.
What is the $\Delta G$ at $300\ K$ for tetramerization of $X$ in liquid phase?

  1. $-52\ kJ/mol$
  2. $-98\ kJ/mol$
  3. $-14.5\ kJ/mol$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

Heat of formation of $2$ moles of ${NH} _{3}(g)$ is $-90kJ$; bond energies of $H-H$ and$N-N$ bonds are $435kJ$ and $390kJ$ ${mol}^{-1}$ respectively. The value of the bond energy of $N\equiv N$ will be:

  1. $-472.5\ kJ$
  2. $-945\ kJ$
  3. $472.5\ kJ$
  4. $945\ kJ$ ${mol}^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\Delta { H } _{ reaction }=\sum { { \left( BE \right)  } _{ reactants } } -\sum { { \left( BE \right)  } _{ products } } $
$-90=x+3\times 435-6\times 390$
$x=945kJ$ ${mol}^{-1}$

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

The standard enthalpy of formation of ${NH} _{3}$ is $-46kJ{ mol }^{ -1 }$. If the enthalpy of formation of ${H} _{2}$ from its atoms is $-436kJ{ mol }^{ -1 }$ and that of ${N} _{2}$ is $-712kJ{ mol }^{ -1 }$, the average bond enthalpy of $N-H$ bond in ${NH} _{3}$ is:

  1. $+1056kJ{ mol }^{ -1 }\quad $
  2. $-1102kJ{ mol }^{ -1 }\quad $
  3. $-964kJ{ mol }^{ -1 }\quad $
  4. $+352kJ{ mol }^{ -1 }\quad $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

${ H } _{ reaction }=\sum { { BE } _{ reactants } } -\sum { { BE } _{ products } } $
$-46=(\cfrac(712)+\cfrac{3}{2}(436))-3x$
$x=+352\ kJ{ mol }^{ -1 } $

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

Calculate $ \Delta { H }^{ o }$ of the reaction:

$ { CH } _{ 2 }={ CH } _{ 2 }+3O=O\longrightarrow 2O=C=O+2H-O-H$

The average bond enthelpies of various bond are:

$Bond \quad \quad \quad\quad \quad C-H\quad \quad O=O\quad \quad C=O\quad \quad O-H\quad \quad C=C$
$Bond\ enthalpy$:     414               499               724           460               619
[kJ/mol]

  1. -364kJ

  2. -564kJ

  3. -964kJ

  4. -1654kJ
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using bond enthalpy method: ΔH = Σ(bonds broken) - Σ(bonds formed). Breaking: 4 C-H (4×414=1656), 1 C=C (619), 3 O=O (3×499=1497). Total broken = 3772. Forming: 4 C=O (4×724=2896), 4 O-H (4×460=1840). Total formed = 4736. ΔH = 3772-4736 = -964 kJ. The reaction is exothermic.

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

$H _{2}+Cl _{2}\rightarrow 2HCl+44\ K.Cal$. Heat of decomposition of $HCl$ is:

  1. $-44\ K.Cal$
  2. $+44\ K.Cal$
  3. $-22\ K.Cal$
  4. $+22\ K.Cal$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$H _2+Cl _2 \longrightarrow 2HCl+44Kcal$

Heat of decomposition of $HCl$ :-
$2HCl \longrightarrow \Delta H _{decomp}=+44 Kcal$
$1 HCl \longrightarrow \Delta H _{decomp}=+22 Kcal$
So, Heat of decomposition of $HCl=+22Kcal$